One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: (a) \(\frac{92}{44} \mathrm{Ru}\) and ${ }^{102} \mathrm{Ru},(\mathbf{b}){ }_{56}^{138} \mathrm{Ba}\( and \){ }_{56}^{139} \mathrm{Ba},(\mathbf{c})$ tin -109 and tin-120.

To calculate the neutron-to-proton ratio, we need to find the number of neutrons in each nuclide and divide it by the number of protons. Protons can be found in the bottom number, while the difference between the top and bottom numbers represents the neutrons. (a) \(\frac{92}{44} \mathrm{Ru}\): Neutrons = 92 - 44 = 48, ratio = \( \frac{48}{44} = 1.09 \) \({ }^{102} \mathrm{Ru}\): Neutrons = 102 - 44 = 58, ratio = \( \frac{58}{44} = 1.32 \) (b) \({ }_{56}^{138} \mathrm{Ba}\): Neutrons = 138 - 56 = 82, ratio = \( \frac{82}{56} = 1.46 \) \({ }_{56}^{139} \mathrm{Ba}\): Neutrons = 139 - 56 = 83, ratio = \( \frac{83}{56} = 1.48 \) (c) Tin-109: Nuclide is \({ }_{50}^{109} \mathrm{Sn}\) Neutrons = 109 - 50 = 59, ratio = \( \frac{59}{50} = 1.18 \) Tin-120: Nuclide is \({ }_{50}^{120} \mathrm{Sn}\) Neutrons = 120 - 50 = 70, ratio = \( \frac{70}{50} = 1.4 \)

Based on the calculated ratios, we can now predict the stability of the nuclides. Remember that lighter elements with a nuclear ratio close to 1 are stable and as the atomic number increases, stable nuclei have higher neutron-to-proton ratios. (a) \(\frac{92}{44} \mathrm{Ru}\) has a ratio of 1.09 while \({ }^{102} \mathrm{Ru}\) has a ratio of 1.32. Since the atomic number is 44 (medium-sized nucleus), a higher neutron-to-proton ratio will make the nucleus more stable. Therefore, \({ }^{102} \mathrm{Ru}\) is stable and \(\frac{92}{44} \mathrm{Ru}\) is radioactive. (b) \({ }_{56}^{138} \mathrm{Ba}\) has a ratio of 1.46 while \({ }_{56}^{139} \mathrm{Ba}\) has a ratio of 1.48. They both have close ratios, but since the atomic number is 56 (medium-sized nucleus), a slightly higher neutron-to-proton ratio will favor stability. Thus, \({ }_{56}^{139} \mathrm{Ba}\) is stable and \({ }_{56}^{138} \mathrm{Ba}\) is radioactive. (c) Tin-109 (\({ }_{50}^{109} \mathrm{Sn}\)) has a ratio of 1.18 while Tin-120 (\({ }_{50}^{120} \mathrm{Sn}\)) has a ratio of 1.4. Since the atomic number is 50 (medium-sized nucleus), a higher neutron-to-proton ratio will favor stability. Therefore, Tin-120 is stable and Tin-109 is radioactive. In conclusion, the radioactive nuclides are \(\frac{92}{44} \mathrm{Ru}\), \({ }_{56}^{138} \mathrm{Ba}\), and Tin-109, while the stable nuclides are \({ }^{102} \mathrm{Ru}\), \({ }_{56}^{139} \mathrm{Ba}\), and Tin-120.